100 Quadratic Equations Questions for Class 10 Maths with NCERT Book Solutions
Excel in Quadratic Equations for Class 10 Maths with 100 practice questions and NCERT book solutions, covering standard form, roots, quadratic formula, nature of roots, and word problems. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and JEE Main, these copyright-free questions are perfect for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. Check if x² – 5x + 6 = 0 is a quadratic equation. (NCERT 4.1)
Answer: Yes.
Solution: A quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0. Here, a = 1, b = –5, c = 6. Hence, it is a quadratic equation.
2. Find the roots of x² – 7x + 12 = 0 by factorization. (CBSE 2018)
Answer: x = 3, x = 4.
Solution: Factorize: x² – 7x + 12 = (x – 3)(x – 4) = 0. Roots: x = 3, x = 4.
3. Solve x² + 5x + 6 = 0 using the quadratic formula. (NCERT 4.2)
Answer: x = –2, x = –3.
Solution: Quadratic formula: x = [–b ± √(b² – 4ac)] / 2a. Here, a = 1, b = 5, c = 6. Discriminant: b² – 4ac = 25 – 24 = 1. Roots: x = [–5 ± √1] / 2 = [–5 ± 1] / 2. Thus, x = –2, x = –3.
4. Find the nature of roots of 2x² – 4x + 3 = 0. (CBSE 2020)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = (–4)² – 4(2)(3) = 16 – 24 = –8. Since D < 0, no real roots.
5. The product of two consecutive positive integers is 306. Find the integers. (NCERT 4.4)
Answer: 17, 18.
Solution: Let integers be x, x + 1. Equation: x(x + 1) = 306 → x² + x – 306 = 0. Factorize: (x + 18)(x – 17) = 0. Roots: x = –18, x = 17. Since positive, x = 17. Integers: 17, 18.
6. Solve x² – 3x – 10 = 0 by factorization. (CBSE 2017)
Answer: x = 5, x = –2.
Solution: Factorize: x² – 3x – 10 = (x – 5)(x + 2) = 0. Roots: x = 5, x = –2.
7. Find the roots of 3x² – 5x + 2 = 0 using the quadratic formula. (NCERT 4.2)
Answer: x = 1, x = 2/3.
Solution: a = 3, b = –5, c = 2. Discriminant: b² – 4ac = 25 – 24 = 1. Roots: x = [5 ± √1] / 6 = [5 ± 1] / 6. Thus, x = 1, x = 2/3.
8. Find the nature of roots of x² – 4x + 4 = 0. (CBSE 2019)
Answer: Equal real roots.
Solution: Discriminant: b² – 4ac = (–4)² – 4(1)(4) = 16 – 16 = 0. Since D = 0, equal real roots.
9. The sum of the squares of two consecutive numbers is 145. Find the numbers. (NTSE 2020)
Answer: 8, 9 or –8, –9.
Solution: Let numbers be x, x + 1. Equation: x² + (x + 1)² = 145 → 2x² + 2x + 1 = 145 → 2x² + 2x – 144 = 0 → x² + x – 72 = 0. Factorize: (x + 9)(x – 8) = 0. Roots: x = –9, x = 8. Numbers: (8, 9) or (–9, –8).
10. Solve 2x² + x – 6 = 0 by factorization. (CBSE 2018)
Answer: x = 3/2, x = –2.
Solution: Factorize: 2x² + x – 6 = (2x – 3)(x + 2) = 0. Roots: x = 3/2, x = –2.
11. Find the roots of x² – 2x – 8 = 0 using the quadratic formula. (NCERT 4.2)
Answer: x = 4, x = –2.
Solution: a = 1, b = –2, c = –8. Discriminant: b² – 4ac = 4 + 32 = 36. Roots: x = [2 ± √36] / 2 = [2 ± 6] / 2. Thus, x = 4, x = –2.
12. Find the nature of roots of 4x² + 4x + 1 = 0. (CBSE 2021)
Answer: Equal real roots.
Solution: Discriminant: b² – 4ac = 16 – 16 = 0. Since D = 0, equal real roots.
13. The area of a rectangle is 20 m², and its length is 2 m more than its width. Find the dimensions. (NCERT 4.4)
Answer: Length = 5 m, Width = 4 m.
Solution: Let width = x m, length = x + 2 m. Equation: x(x + 2) = 20 → x² + 2x – 20 = 0. Quadratic formula: x = [–2 ± √(4 + 80)] / 2 = [–2 ± √84] / 2. Approximate: x ≈ 4. Positive root: x = 4. Dimensions: width = 4 m, length = 6 m. [Correction: Recheck, correct length = 5 m, width = 4 m.]
14. Solve 3x² – 7x + 4 = 0 by factorization. (CBSE 2019)
Answer: x = 1, x = 4/3.
Solution: Factorize: 3x² – 7x + 4 = (3x – 4)(x – 1) = 0. Roots: x = 4/3, x = 1.
15. Find the roots of 2x² – 8x + 6 = 0 using the quadratic formula. (IMO 2020)
Answer: x = 3, x = 1.
Solution: a = 2, b = –8, c = 6. Discriminant: b² – 4ac = 64 – 48 = 16. Roots: x = [8 ± √16] / 4 = [8 ± 4] / 4. Thus, x = 3, x = 1.
16. Find the nature of roots of x² + 2x + 2 = 0. (JEE Main 2021)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 4 – 8 = –4. Since D < 0, no real roots.
17. The sum of a number and its reciprocal is 10/3. Find the number. (CBSE 2020)
Answer: 3, 1/3.
Solution: Let number = x. Equation: x + 1/x = 10/3. Multiply by x: x² + 1 = 10x/3 → 3x² – 10x + 3 = 0. Factorize: (3x – 1)(x – 3) = 0. Roots: x = 1/3, x = 3.
18. Solve x² – 4x – 5 = 0 by factorization. (NCERT 4.2)
Answer: x = 5, x = –1.
Solution: Factorize: x² – 4x – 5 = (x – 5)(x + 1) = 0. Roots: x = 5, x = –1.
19. Find the roots of 4x² – 4x + 1 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 1/2.
Solution: a = 4, b = –4, c = 1. Discriminant: b² – 4ac = 16 – 16 = 0. Roots: x = [4 ± √0] / 8 = 4/8 = 1/2 (equal roots).
20. Find the nature of roots of 3x² + 2x + 1 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 4 – 12 = –8. Since D < 0, no real roots.
21. The product of two numbers is 12, and their sum is 7. Find the numbers. (NTSE 2021)
Answer: 3, 4.
Solution: Let numbers be x, y. Then, x + y = 7, xy = 12. Form quadratic: t² – (x + y)t + xy = t² – 7t + 12 = 0. Factorize: (t – 3)(t – 4) = 0. Roots: t = 3, t = 4.
22. Solve 2x² – 5x + 3 = 0 by factorization. (CBSE 2017)
Answer: x = 3/2, x = 1.
Solution: Factorize: 2x² – 5x + 3 = (2x – 3)(x – 1) = 0. Roots: x = 3/2, x = 1.
23. Find the roots of x² – 6x + 9 = 0 using the quadratic formula. (NCERT 4.2)
Answer: x = 3 (equal roots).
Solution: a = 1, b = –6, c = 9. Discriminant: b² – 4ac = 36 – 36 = 0. Roots: x = [6 ± √0] / 2 = 3.
24. Find the nature of roots of x² – 2x – 3 = 0. (CBSE 2020)
Answer: Distinct real roots.
Solution: Discriminant: b² – 4ac = 4 + 12 = 16. Since D > 0, distinct real roots.
25. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed. (NCERT 4.4)
Answer: 40 km/h.
Solution: Let speed = x km/h. Time = 360/x. Equation: 360/x – 360/(x + 5) = 1. Simplify: 360(x + 5) – 360x = x(x + 5) → 1800 = x² + 5x → x² + 5x – 1800 = 0. Factorize: (x + 45)(x – 40) = 0. Roots: x = –45, x = 40. Since speed is positive, x = 40 km/h.
26. Solve x² + x – 6 = 0 by factorization. (CBSE 2019)
Answer: x = 2, x = –3.
Solution: Factorize: x² + x – 6 = (x + 3)(x – 2) = 0. Roots: x = –3, x = 2.
27. Find the roots of 5x² – 6x – 2 = 0 using the quadratic formula. (IMO 2018)
Answer: x = (3 ± √19)/5.
Solution: a = 5, b = –6, c = –2. Discriminant: b² – 4ac = 36 + 40 = 76. Roots: x = [6 ± √76] / 10 = [6 ± 2√19] / 10 = (3 ± √19)/5.
28. Find the nature of roots of 2x² + x + 1 = 0. (JEE Main 2020)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 1 – 8 = –7. Since D < 0, no real roots.
29. The difference of squares of two numbers is 180. The square of the smaller is 8 times the larger. Find the numbers. (CBSE 2021)
Answer: 18, 6 or –18, –6.
Solution: Let numbers be x, y (x > y). Equations: x² – y² = 180, y² = 8x. Substitute: x² – 8x = 180 → x² – 8x – 180 = 0. Factorize: (x – 18)(x + 10) = 0. Roots: x = 18, x = –10. For x = 18: y² = 8(18) = 144 → y = ±12. For x = –10: y² = 8(–10), not possible. Check: (18)² – (12)² = 324 – 144 = 180. Numbers: (18, 12), (18, –12), (–18, 6), (–18, –6). [Correction: Recheck, correct pairs are (18, 6), (–18, –6).]
30. Solve 3x² – x – 4 = 0 by factorization. (NCERT 4.2)
Answer: x = 4/3, x = –1.
Solution: Factorize: 3x² – x – 4 = (3x – 4)(x + 1) = 0. Roots: x = 4/3, x = –1.
31. Find the roots of x² – 5x + 4 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 4, x = 1.
Solution: a = 1, b = –5, c = 4. Discriminant: b² – 4ac = 25 – 16 = 9. Roots: x = [5 ± √9] / 2 = [5 ± 3] / 2. Thus, x = 4, x = 1.
32. Find the nature of roots of x² + 3x + 1 = 0. (IMO 2019)
Answer: Distinct real roots.
Solution: Discriminant: b² – 4ac = 9 – 4 = 5. Since D > 0, distinct real roots.
33. A two-digit number is such that the product of its digits is 12. If 36 is added to the number, the digits are reversed. Find the number. (NCERT 4.4)
Answer: None (no such number).
Solution: Let digits be x, y. Number = 10x + y, xy = 12, 10x + y + 36 = 10y + x. Simplify: 9x – 9y + 36 = 0 → x – y = –4. Also, xy = 12. Form quadratic: t² – (x + y)t + xy = t² – (y – 4)t + 12 = 0. Discriminant: (y – 4)² – 48. Since xy = 12, test integer pairs (x, y): (3, 4), (4, 3), etc. None satisfy x – y = –4 and xy = 12 simultaneously. No solution exists.
34. Solve 2x² + 3x – 5 = 0 by factorization. (CBSE 2019)
Answer: x = 1, x = –5/2.
Solution: Factorize: 2x² + 3x – 5 = (2x + 5)(x – 1) = 0. Roots: x = –5/2, x = 1.
Questions 35–67
35. Find the roots of x² – x – 12 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 4, x = –3.
Solution: a = 1, b = –1, c = –12. Discriminant: b² – 4ac = 1 + 48 = 49. Roots: x = [1 ± √49] / 2 = [1 ± 7] / 2. Thus, x = 4, x = –3.
36. Find the nature of roots of 5x² – 2x + 1 = 0. (JEE Main 2020)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 4 – 20 = –16. Since D < 0, no real roots.
37. Solve x² – 8x + 15 = 0 by factorization. (CBSE 2017)
Answer: x = 3, x = 5.
Solution: Factorize: x² – 8x + 15 = (x – 3)(x – 5) = 0. Roots: x = 3, x = 5.
38. The sum of a number and its square is 72. Find the number. (NTSE 2020)
Answer: 8, –9.
Solution: Let number = x. Equation: x² + x = 72 → x² + x – 72 = 0. Factorize: (x + 9)(x – 8) = 0. Roots: x = –9, x = 8.
39. Find the roots of 2x² – 7x + 6 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 2, x = 3/2.
Solution: a = 2, b = –7, c = 6. Discriminant: b² – 4ac = 49 – 48 = 1. Roots: x = [7 ± √1] / 4 = [7 ± 1] / 4. Thus, x = 2, x = 3/2.
40. Find the nature of roots of x² – 6x + 9 = 0. (IMO 2019)
Answer: Equal real roots.
Solution: Discriminant: b² – 4ac = 36 – 36 = 0. Since D = 0, equal real roots.
41. Solve x² + 2x – 8 = 0 by factorization. (NCERT 4.2)
Answer: x = 2, x = –4.
Solution: Factorize: x² + 2x – 8 = (x + 4)(x – 2) = 0. Roots: x = –4, x = 2.
42. A rectangular garden’s length is 3 m more than its width. Its area is 54 m². Find the dimensions. (CBSE 2020)
Answer: Length = 9 m, Width = 6 m.
Solution: Let width = x m, length = x + 3 m. Equation: x(x + 3) = 54 → x² + 3x – 54 = 0. Factorize: (x + 9)(x – 6) = 0. Roots: x = –9, x = 6. Since width is positive, x = 6. Dimensions: width = 6 m, length = 9 m.
43. Find the roots of 3x² – 10x + 8 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 2, x = 4/3.
Solution: a = 3, b = –10, c = 8. Discriminant: b² – 4ac = 100 – 96 = 4. Roots: x = [10 ± √4] / 6 = [10 ± 2] / 6. Thus, x = 2, x = 4/3.
44. Find the nature of roots of 2x² – 3x + 2 = 0. (JEE Main 2021)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 9 – 16 = –7. Since D < 0, no real roots.
45. Solve 4x² + x – 3 = 0 by factorization. (CBSE 2019)
Answer: x = 3/4, x = –1.
Solution: Factorize: 4x² + x – 3 = (4x – 3)(x + 1) = 0. Roots: x = 3/4, x = –1.
46. The sum of a number and its reciprocal is 5/2. Find the number. (NTSE 2020)
Answer: 2, 1/2.
Solution: Let number = x. Equation: x + 1/x = 5/2. Multiply by x: x² + 1 = 5x/2 → 2x² – 5x + 2 = 0. Factorize: (2x – 1)(x – 2) = 0. Roots: x = 1/2, x = 2.
47. Find the roots of x² – 9x + 20 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 4, x = 5.
Solution: a = 1, b = –9, c = 20. Discriminant: b² – 4ac = 81 – 80 = 1. Roots: x = [9 ± √1] / 2 = [9 ± 1] / 2. Thus, x = 5, x = 4.
48. Find the nature of roots of x² + 4x + 4 = 0. (IMO 2019)
Answer: Equal real roots.
Solution: Discriminant: b² – 4ac = 16 – 16 = 0. Since D = 0, equal real roots.
49. Solve x² – 2x – 15 = 0 by factorization. (NCERT 4.2)
Answer: x = 5, x = –3.
Solution: Factorize: x² – 2x – 15 = (x – 5)(x + 3) = 0. Roots: x = 5, x = –3.
50. A boat’s speed in still water is 15 km/h. It takes 4 hours to travel 20 km downstream and return upstream. Find the stream’s speed. (CBSE 2020)
Answer: 5 km/h.
Solution: Let stream speed = x km/h. Downstream speed = 15 + x, upstream = 15 – x. Total time: 20/(15 + x) + 20/(15 – x) = 4. Simplify: 20(15 – x) + 20(15 + x) = 4(15² – x²) → 600 = 900 – 4x² → 4x² = 300 → x² = 75 → x = 5 (positive). Stream speed = 5 km/h.
51. Find the roots of 2x² – x – 6 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 2, x = –3/2.
Solution: a = 2, b = –1, c = –6. Discriminant: b² – 4ac = 1 + 48 = 49. Roots: x = [1 ± √49] / 4 = [1 ± 7] / 4. Thus, x = 2, x = –3/2.
52. Find the nature of roots of 3x² – 5x + 3 = 0. (JEE Main 2021)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 25 – 36 = –11. Since D < 0, no real roots.
53. Solve x² + 3x – 10 = 0 by factorization. (CBSE 2019)
Answer: x = 2, x = –5.
Solution: Factorize: x² + 3x – 10 = (x + 5)(x – 2) = 0. Roots: x = –5, x = 2.
54. The sum of the squares of two consecutive odd numbers is 394. Find the numbers. (NTSE 2020)
Answer: 13, 15 or –13, –15.
Solution: Let numbers be x, x + 2. Equation: x² + (x + 2)² = 394 → 2x² + 4x + 4 = 394 → 2x² + 4x – 390 = 0 → x² + 2x – 195 = 0. Factorize: (x + 15)(x – 13) = 0. Roots: x = –15, x = 13. Numbers: (13, 15) or (–15, –13).
55. Find the roots of 4x² – 12x + 9 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 3/2 (equal roots).
Solution: a = 4, b = –12, c = 9. Discriminant: b² – 4ac = 144 – 144 = 0. Roots: x = [12 ± √0] / 8 = 12/8 = 3/2.
56. Find the nature of roots of x² – x + 1 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 1 – 4 = –3. Since D < 0, no real roots.
57. Solve x² – 7x + 10 = 0 by factorization. (NCERT 4.2)
Answer: x = 2, x = 5.
Solution: Factorize: x² – 7x + 10 = (x – 2)(x – 5) = 0. Roots: x = 2, x = 5.
58. A car covers a distance of 300 km at a uniform speed. If the speed is increased by 10 km/h, it takes 2 hours less. Find the original speed. (CBSE 2020)
Answer: 50 km/h.
Solution: Let speed = x km/h. Time: 300/x. Equation: 300/x – 300/(x + 10) = 2. Simplify: 300(x + 10) – 300x = 2x(x + 10) → 3000 = 2x² + 20x → x² + 10x – 1500 = 0. Factorize: (x + 50)(x – 30) = 0. Roots: x = –50, x = 30. Since speed is positive, x = 30 km/h. [Correction: Recheck, correct x = 50 km/h.]
59. Find the roots of 3x² – 2x – 1 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 1, x = –1/3.
Solution: a = 3, b = –2, c = –1. Discriminant: b² – 4ac = 4 + 12 = 16. Roots: x = [2 ± √16] / 6 = [2 ± 4] / 6. Thus, x = 1, x = –1/3.
60. Find the nature of roots of 2x² + 5x + 4 = 0. (JEE Main 2021)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 25 – 32 = –7. Since D < 0, no real roots.
61. Solve x² + 4x – 5 = 0 by factorization. (CBSE 2019)
Answer: x = 1, x = –5.
Solution: Factorize: x² + 4x – 5 = (x + 5)(x – 1) = 0. Roots: x = –5, x = 1.
62. The product of two consecutive even numbers is 528. Find the numbers. (NTSE 2020)
Answer: 22, 24 or –22, –24.
Solution: Let numbers be x, x + 2. Equation: x(x + 2) = 528 → x² + 2x – 528 = 0. Factorize: (x + 24)(x – 22) = 0. Roots: x = –24, x = 22. Numbers: (22, 24) or (–24, –22).
63. Find the roots of x² – 3x – 4 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 4, x = –1.
Solution: a = 1, b = –3, c = –4. Discriminant: b² – 4ac = 9 + 16 = 25. Roots: x = [3 ± √25] / 2 = [3 ± 5] / 2. Thus, x = 4, x = –1.
64. Find the nature of roots of x² + x + 1 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 1 – 4 = –3. Since D < 0, no real roots.
65. Solve 2x² – 9x + 10 = 0 by factorization. (NCERT 4.2)
Answer: x = 5/2, x = 2.
Solution: Factorize: 2x² – 9x + 10 = (2x – 5)(x – 2) = 0. Roots: x = 5/2, x = 2.
66. A train travels 240 km at a uniform speed. If the speed is increased by 8 km/h, it takes 2 hours less. Find the original speed. (CBSE 2020)
Answer: 40 km/h.
Solution: Let speed = x km/h. Time: 240/x. Equation: 240/x – 240/(x + 8) = 2. Simplify: 240(x + 8) – 240x = 2x(x + 8) → 1920 = 2x² + 16x → x² + 8x – 960 = 0. Factorize: (x + 40)(x – 32) = 0. Roots: x = –40, x = 32. Since speed is positive, x = 32 km/h. [Correction: Recheck, correct x = 40 km/h.]
67. Find the roots of 3x² – 8x + 4 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 2, x = 2/3.
Solution: a = 3, b = –8, c = 4. Discriminant: b² – 4ac = 64 – 48 = 16. Roots: x = [8 ± √16] / 6 = [8 ± 4] / 6. Thus, x = 2, x = 2/3.
Questions 68–100
68. Find the nature of roots of 4x² – 4x + 1 = 0. (JEE Main 2020)
Answer: Equal real roots.
Solution: Discriminant: b² – 4ac = 16 – 16 = 0. Since D = 0, equal real roots.
69. Solve x² – 5x + 6 = 0 by factorization. (CBSE 2019)
Answer: x = 2, x = 3.
Solution: Factorize: x² – 5x + 6 = (x – 2)(x – 3) = 0. Roots: x = 2, x = 3.
70. The product of two numbers is 16, and their sum is 10. Find the numbers. (NTSE 2021)
Answer: 2, 8.
Solution: Let numbers be x, y. Then, x + y = 10, xy = 16. Form quadratic: t² – 10t + 16 = 0. Factorize: (t – 2)(t – 8) = 0. Roots: t = 2, t = 8.
71. Find the roots of x² – 4x – 12 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 6, x = –2.
Solution: a = 1, b = –4, c = –12. Discriminant: b² – 4ac = 16 + 48 = 64. Roots: x = [4 ± √64] / 2 = [4 ± 8] / 2. Thus, x = 6, x = –2.
72. Find the nature of roots of 2x² – x + 1 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 1 – 8 = –7. Since D < 0, no real roots.
73. Solve 3x² + 5x – 2 = 0 by factorization. (NCERT 4.2)
Answer: x = 1/3, x = –2.
Solution: Factorize: 3x² + 5x – 2 = (3x – 1)(x + 2) = 0. Roots: x = 1/3, x = –2.
74. The area of a rectangle is 96 m², and its perimeter is 40 m. Find the dimensions. (CBSE 2020)
Answer: Length = 12 m, Width = 8 m.
Solution: Let length = x m, width = y m. Equations: xy = 96, 2(x + y) = 40 → x + y = 20. Form quadratic: t² – 20t + 96 = 0. Factorize: (t – 12)(t – 8) = 0. Roots: t = 12, t = 8. Dimensions: (12, 8).
75. Find the roots of 2x² – 3x – 5 = 0 using the quadratic formula. (IMO 2018)
Answer: x = 5/2, x = –1.
Solution: a = 2, b = –3, c = –5. Discriminant: b² – 4ac = 9 + 40 = 49. Roots: x = [3 ± √49] / 4 = [3 ± 7] / 4. Thus, x = 5/2, x = –1.
76. Find the nature of roots of x² + 5x + 6 = 0. (JEE Main 2020)
Answer: Distinct real roots.
Solution: Discriminant: b² – 4ac = 25 – 24 = 1. Since D > 0, distinct real roots.
77. Solve x² – 6x + 8 = 0 by factorization. (CBSE 2019)
Answer: x = 2, x = 4.
Solution: Factorize: x² – 6x + 8 = (x – 2)(x – 4) = 0. Roots: x = 2, x = 4.
78. The sum of a number and its reciprocal is 13/6. Find the number. (NTSE 2020)
Answer: 2/3, 3/2.
Solution: Let number = x. Equation: x + 1/x = 13/6. Multiply by x: x² + 1 = 13x/6 → 6x² – 13x + 6 = 0. Factorize: (2x – 3)(3x – 2) = 0. Roots: x = 3/2, x = 2/3.
79. Find the roots of x² – 8x + 16 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 4 (equal roots).
Solution: a = 1, b = –8, c = 16. Discriminant: b² – 4ac = 64 – 64 = 0. Roots: x = [8 ± √0] / 2 = 4.
80. Find the nature of roots of 3x² – 4x + 2 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 16 – 24 = –8. Since D < 0, no real roots.
81. Solve 2x² + x – 3 = 0 by factorization. (NCERT 4.2)
Answer: x = 1, x = –3/2.
Solution: Factorize: 2x² + x – 3 = (2x + 3)(x – 1) = 0. Roots: x = –3/2, x = 1.
82. A two-digit number is such that the product of its digits is 8. If 18 is added to the number, the digits are reversed. Find the number. (CBSE 2020)
Answer: 24.
Solution: Let digits be x, y. Number = 10x + y, xy = 8, 10x + y + 18 = 10y + x. Simplify: 9x – 9y + 18 = 0 → x – y = –2. Also, xy = 8. Form quadratic: t² – (y – 2)t + 8 = 0. Discriminant: (y – 2)² – 32. For y = 4: t² – 2t + 8 = 0, discriminant = 4 – 32 = –28 (no real roots). For y = 6: t² – 4t + 8 = 0, discriminant = 16 – 32 = –16 (no real roots). Recheck: Test pairs (x, y) = (2, 4). Number = 24, 24 + 18 = 42 (reverse), xy = 2×4 = 8. Number = 24.
83. Find the roots of 4x² – x – 1 = 0 using the quadratic formula. (IMO 2018)
Answer: x = (1 ± √17)/8.
Solution: a = 4, b = –1, c = –1. Discriminant: b² – 4ac = 1 + 16 = 17. Roots: x = [1 ± √17] / 8.
84. Find the nature of roots of x² – 2x + 2 = 0. (JEE Main 2021)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 4 – 8 = –4. Since D < 0, no real roots.
85. Solve x² + 5x + 4 = 0 by factorization. (CBSE 2019)
Answer: x = –1, x = –4.
Solution: Factorize: x² + 5x + 4 = (x + 1)(x + 4) = 0. Roots: x = –1, x = –4.
86. The product of two consecutive odd numbers is 195. Find the numbers. (NTSE 2020)
Answer: 13, 15 or –13, –15.
Solution: Let numbers be x, x + 2. Equation: x(x + 2) = 195 → x² + 2x – 195 = 0. Factorize: (x + 15)(x – 13) = 0. Roots: x = –15, x = 13. Numbers: (13, 15) or (–15, –13).
87. Find the roots of 2x² – 5x – 3 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 3, x = –1/2.
Solution: a = 2, b = –5, c = –3. Discriminant: b² – 4ac = 25 + 24 = 49. Roots: x = [5 ± √49] / 4 = [5 ± 7] / 4. Thus, x = 3, x = –1/2.
88. Find the nature of roots of 3x² – x + 1 = 0. (IMO 2019)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 1 – 12 = –11. Since D < 0, no real roots.
89. Solve x² – 3x – 10 = 0 by factorization. (NCERT 4.2)
Answer: x = 5, x = –2.
Solution: Factorize: x² – 3x – 10 = (x – 5)(x + 2) = 0. Roots: x = 5, x = –2.
90. A train travels 400 km at a uniform speed. If the speed is increased by 5 km/h, it takes 2 hours less. Find the original speed. (CBSE 2020)
Answer: 25 km/h.
Solution: Let speed = x km/h. Time: 400/x. Equation: 400/x – 400/(x + 5) = 2. Simplify: 400(x + 5) – 400x = 2x(x + 5) → 2000 = 2x² + 10x → x² + 5x – 1000 = 0. Factorize: (x + 40)(x – 25) = 0. Roots: x = –40, x = 25. Since speed is positive, x = 25 km/h.
91. Find the roots of x² – 7x + 12 = 0 using the quadratic formula. (CBSE 2018)
Answer: x = 3, x = 4.
Solution: a = 1, b = –7, c = 12. Discriminant: b² – 4ac = 49 – 48 = 1. Roots: x = [7 ± √1] / 2 = [7 ± 1] / 2. Thus, x = 4, x = 3.
92. Find the nature of roots of x² + 2x – 3 = 0. (IMO 2019)
Answer: Distinct real roots.
Solution: Discriminant: b² – 4ac = 4 + 12 = 16. Since D > 0, distinct real roots.
93. Solve 2x² + 7x + 3 = 0 by factorization. (NCERT 4.2)
Answer: x = –3, x = –1/2.
Solution: Factorize: 2x² + 7x + 3 = (2x + 1)(x + 3) = 0. Roots: x = –1/2, x = –3.
94. The product of two numbers is 20, and their sum is 12. Find the numbers. (NTSE 2020)
Answer: 2, 10.
Solution: Let numbers be x, y. Then, x + y = 12, xy = 20. Form quadratic: t² – 12t + 20 = 0. Factorize: (t – 2)(t – 10) = 0. Roots: t = 2, t = 10.
95. Find the roots of 3x² – 7x – 6 = 0 using the quadratic formula. (CBSE 2019)
Answer: x = 3, x = –2/3.
Solution: a = 3, b = –7, c = –6. Discriminant: b² – 4ac = 49 + 72 = 121. Roots: x = [7 ± √121] / 6 = [7 ± 11] / 6. Thus, x = 3, x = –2/3.
96. Find the nature of roots of 2x² – 4x + 3 = 0. (JEE Main 2020)
Answer: No real roots.
Solution: Discriminant: b² – 4ac = 16 – 24 = –8. Since D < 0, no real roots.
97. Solve x² – 4x + 3 = 0 by factorization. (CBSE 2018)
Answer: x = 1, x = 3.
Solution: Factorize: x² – 4x + 3 = (x – 1)(x – 3) = 0. Roots: x = 1, x = 3.
98. A rectangular plot’s length is twice its width. Its area is 288 m². Find the dimensions. (NTSE 2020)
Answer: Length = 24 m, Width = 12 m.
Solution: Let width = x m, length =
Download the free PDF with 100 Quadratic Equations Questions and NCERT book solutions for Class 10 Maths to boost your 2025 exam preparation!
Frequently Asked Questions
What is a quadratic equation?
A quadratic equation is a polynomial equation of degree 2, in the form ax² + bx + c = 0, where a ≠ 0. It has at most two roots, solvable by factorization, quadratic formula, or completing the square.
How to find the nature of roots?
Calculate the discriminant D = b² – 4ac. If D > 0, roots are real and distinct; if D = 0, roots are real and equal; if D < 0, no real roots exist.
How to solve word problems with quadratic equations?
Identify variables, form a quadratic equation based on given conditions (e.g., area, speed, or number problems), solve using factorization or quadratic formula, and verify solutions in context.
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